464 Higher Engineering Mathematics
For line 2,wherex 0 = 0 .2andh= 0 .2:
y 1 =y 0 +h(y′), from equation( 2 )
= 1 +( 0. 2 )(− 1 )=0.8
and (y′) 0 = 2 x 0 −y 0 = 2 ( 0. 2 )− 0. 8 =−0.4
For line 3,wherex 0 = 0 .4:
y 1 =y 0 +h(y′) 0
= 0. 8 +( 0. 2 )(− 0. 4 )=0.72
and (y′) 0 = 2 x 0 −y 0 = 2 ( 0. 4 )− 0. 72 =0.08
For line 4,wherex 0 = 0 .6:
y 1 =y 0 +h(y′) 0
= 0. 72 +( 0. 2 )( 0. 08 )=0.736
and (y′) 0 = 2 x 0 −y 0 = 2 ( 0. 6 )− 0. 736 =0.464
For line 5,wherex 0 = 0 .8:
y 1 =y 0 +h(y′) 0
= 0. 736 +( 0. 2 )( 0. 464 )=0.8288
and(y′) 0 = 2 x 0 −y 0 = 2 ( 0. 8 )− 0. 8288 =0.7712
For line 6,wherex 0 = 1 .0:
y 1 =y 0 +h(y′) 0
= 0. 8288 +( 0. 2 )( 0. 7712 )=0.98304
As the range is 0 to 1.0,(y′) 0 in line 6 is not needed.
y
x
1.0
0.5
0 0.2 0.4 0.6 0.8 1.0
Figure 49.7
A graph of the solution of
dy
dx
+y= 2 x, with initial
conditionsx=0andy=1 is shown in Fig. 49.7.
Problem 3.
(a) Obtain a numerical solution, using
Euler’s method, of the differential equation
dy
dx
=y−x, with the initial conditions that at
x=0,y=2, for the rangex= 0 ( 0. 1 ) 0 .5. Draw
the graph of the solution.
(b) By an analytical method (using the integrating
factor method of Chapter 48), the solution of
the above differential equation is given by
y=x+ 1 +ex.
Determine the percentage error atx= 0. 3
(a)
dy
dx
=y′=y−x.
If initiallyx 0 =0andy 0 =2,
then(y′) 0 =y 0 −x 0 = 2 − 0 = 2.
Hence line 1 of Table 49.3 is completed.
For line 2,wherex 0 = 0 .1:
y 1 =y 0 +h(y′) 0 , from equation( 2 ),
= 2 +( 0. 1 )( 2 )=2.2
and(y′) 0 =y 0 −x 0
= 2. 2 − 0. 1 =2.1
For line 3,wherex 0 = 0 .2:
y 1 =y 0 +h(y′) 0
= 2. 2 +( 0. 1 )( 2. 1 )=2.41
and(y′) 0 =y 0 −x 0 = 2. 41 − 0. 2 =2.21
Table 49.3
x 0 y 0 (y′) 0
- 0 2 2
- 0.1 2.2 2.1
- 0.2 2.41 2.21
- 0.3 2.631 2.331
- 0.4 2.8641 2.4641
- 0.5 3.11051